No Integer allowed

In this solution, $[x]$ will denote the greatest integer function (the greatest integer less than or equal to $x$ )

Note that all positive real numbers can be uniquely represented by $x=\sqrt{a^2+k}\qquad a\in\mathbb{Z}, 0\le k < 2a+1$ because $a^2+k$ also uniquely represents all real numbers.

Plugging this expression in for $x$ gives $9+[\sqrt{a^2+k}]^2+[2\sqrt{a^2+k}]^2=[(\sqrt{a^2+k})^2]+[(2\sqrt{a^2+k})^2]$ Now we simplify: $9+[\sqrt{a^2+k}]^2+[\sqrt{4a^2+4k}]^2=[a^2+k]+[4a^2+4k]$

Note that since $a\le \sqrt{a^2+k} < a+1$ we must have $[\sqrt{a^2+k}]=a$ . Also, note that $[a^2+k]=a^2+[k]$ and $[4a^2+4k]=4a^2+[4k]$ which simplifies matters to $9+[\sqrt{4a^2+4k}]^2=4a^2+[k]+[4k]$ after cancelling $a^2$ from both sides.

Finally, we notice that when $0\le k\le a+\dfrac{1}{4}$ , then $2a\le \sqrt{4a^2+4k} < 2a+1$ and when $a+\dfrac{1}{4}\le k < 2a+1$ then $2a+1\le \sqrt{4a^2+4k} < 2a+2$ so we have two cases:

Case 1: $0\le k< a+\dfrac{1}{4}$

In this case, $[\sqrt{4a^2+4k}]=2a$ so our equation simplifies to $9=[k]+[4k]$ Note that the RHS is an increasing sequence, and that $1+\dfrac{3}{4} \le k < 2$ gives $RHS=8$ while $k=2$ gives $RHS=10$ , so this case is impossible to satisfy.

Case 2: $a+\dfrac{1}{4}\le k<2a+1$

In this case, $[\sqrt{4a^2+4k}]=2a+1$ so our equation turns into $10+4a=[k]+[4k]$ after simplification. To maximize $x$ , we try $a=9$ . Then our restriction is $9+\dfrac{1}{4}\le k < 19$ and $[k]+[4k]=46$ which is satisfied when $9+\dfrac{1}{4}\le k < 9+\dfrac{1}{2}$ which fits our restriction.

Thus, $x=\sqrt{9^2+9+\dfrac{1}{2}-\varepsilon}=\sqrt{\dfrac{181}{2}-\varepsilon}$ where $\varepsilon > 0$ is a very small number. No matter, we get that $[100x] = \boxed{951}$ with a calculator and we are done.

You can substitute $x$ for $x=N+0.5$ , where $N$ is an integer. The floor brackets that do not have a remaining constant term inside them can be removed, and $N$ comes out to be $9$ . As a result, $x=9+0.5=9.5$ . However, this is not the maximum solution...

Now by testing $9.55$ , $9.6$ and $9.7$ , they do not work. This means that $9.5<x<9.55$ . By testing $9.51$ , the correct answer of $951$ can be derived.